elementary statics and dynamics

8/13/2019 Elementary Statics and Dynamics

1/15

CBSE CLASS XII MATHSElementary Statics And Dynamics

Two mark questions with answers

Q1.To a man walking at the rate of 8 km h-1, rain appears to fall vertically

downwards. find the actual direction of the rain if its actual velocity is 82 km h-1.

Ans1.Let the man e moving along !" at the rate of 8 km h-1and let the rain

appear to fall along !#$. Let the rain make angle with !#. %ain is falling with avelocity of 82 kmh-1.

82 kmh-1is the resultant of 8 kmh-1and &%'.%esolving along !"82.cos(2)* + 8cos+ &%'.cos2)*o

82sin 8 &%'velocity rain /.r.t man0sin 12 3o.

Q2.Two forces of magnitudes 24 and 4 act an a point and the angle etween them

is 5*o. 6ind the magnitude and the direction of their resultant.

Ans2.%esultant for the forces 7 and inclined at an angle is

%esultant (22+ 2+ 2x2x cos5*9 ( + 15 + 8 2) 4:f the resultant makes angle with 7;irection of resultant is tan sin(7 + costan (2xsin5*o(2 + xcos 5*o0


2/15


3/15

these stringsF

Ans".Let T and T$ e the two tensions. =ince the mass is in equilirium, the

resultant of the tensions must e equal and opposite to the 1* kg w.t.

%esolve the 1*kg w.t. along the lines of action of the tension i.e., along lines makingangles


4/15

/e have, T and % .

Therefore,

tan

.

Q11.=how that the path of a particle moving in space with constant acceleration is aparaolaF

Ans11.Ccceleration is constant.

constant a (say.

:ntegrating, we get aI + .

Cgain integrating, we get y , which is a paraola.

Q12.C truck is moving along a level road at the rate of )2 km per hour. :n whatdirection must a ullet e shot from it with a velocity of


5/15

&G.

&G K


6/15

Ans1.=uppose the forces 7 and act at an angle and let their resultant e %.Then

%2 72+ 2+ 27 cos .....(i:t is given that % is also the resultant of the forces 7 + = and - = acting at the

same angle. Therefore,

%2 (7 + =2+ ( - =2+ 2(7 + =( - = cos ....(ii6rom (i and (ii, we get

72+ 2+ 27 cos (7 + =2+ ( - =2+ 2(7 + = ( - = cos * 2=2+ 2=(7 - + 2=( - 7 cos - 2=2cos * 2=K(= + 7 - - (= + 7 - cos (= + 7 - cos (= + 7 -

cos 1 ...= - 7 ... (= + 7 - *07utting cos 1 in (i, we otain,%2 72+ 2+ 27 (7 + 2

% 7 + >ence, the magnitude of the resultant of 7 and is 7 + .

Q2.Two forces each of magnitude 7 act at a point. 6ind the angle etween tham ifthe magnitude of their resultant is equal to (1 7 and (2 72.

Ans2.Let e the angle etween the two equal forces each of magnitude 7. Then,their resultant % is given y

% 27cos(2 .......(sustituting the values in % (72+ q2+ 2pq cos(1 :f the resultant of the equal forces is 7, then,

% 27 cos(27 27 cos(2 ...% 701 2 cos(2cos(2 12 cos(2 cos5*92 5*9 12*9>ence, the angle etween the forces is 12*9.

(2 :f the resultant of the equal forces is 72, then,

% 27 cos(272 27 cos(2cos(2 12 cos-1(1 2cos-1(1 cos-1K2x(12- 1 ...2cos-1I cos-1(2I2- 10 cos-1(-)8>ence, the angle etween the forces is cos-1(-)8.

Q3./ith two forces acting at a point, the maIimum effect is otained when their

resultant is 4ewtons. :f they act at right angles, their resultant is < 4ewtons. 6indthe forces.

Ans3.Let the two forces e of magnitudes 7 and 4ewtons. /e know that resultant

(% of two forces (7 and is maIimum when cosis maIimum. Gut maIimum valueof cos 1, when * then,%2 72+ 2+ 27.1

%2 (7 + 2

% 7 +


7/15


8/15


9/15

cos N(


10/15

< sin (7 + cos< sin 7 + (22- 72270 ....using (i0< sin 7 + (22- 72270< sin (72+ 222sin (72+ 2227 < ......(ii4ow, cos2+ sin2 1(22- 722702+ (72+ 2227


11/15

&72- u2 2gh -------(2

u2- 2gh0 - u2 2gh

u2- 8gh - u2 2gh


12/15

Ans3.Let the particle e proJected from C and let it returns to the ground aG. Let 7

e the position of the particle at some instant moving with velocity u at angle withthe horiDontal. Let after time t. The particles reaches where its velocity is v O

makes an angle (?*P - to the horiDontal component of velocity constantthroughout the motion.

>oriDontal component of velocity of 7 horiDontal component of velocity at ucos vcos (?*P - ucos vsin....................(iClso, for vertical of velocity, using equation

v u - gt, for upward motioninitial velocity is usinand final velocity eing - vcos.-vcos usin- gtgt usin+ vcos.................(ii7utting the value of v from (i in (iigt usin+ ucossinxcosgt u (sin2+ cos2sin usingt u cosect ug cosec>ence, the particle will e at right angle to the previous direction after time ug

cosec.

Q4.:f v1O v2e the velocities at the ends of a focal chord of a proJectile path O u e

the velocity at the verteI of the path. Then show that 1v12+ 1v2

2 1u2.

Ans4.

Let in figure 7 e the given focal chord of the paraolic traJectory followed y aparticle proJected from !. /e know that the tangents drawn at the end points of a

focal chord intersect at right angles. ... :f v1is the velocity of the particle at 7 makingan angle with the horiDontal 7I$. Then velocity v2of the particle at will make anangle (?*P - to the horiDontal. Let C e the verteI of the paraola where velocity

of the particle is u, directed horiDontally.... >oriDontal component of velocity remains constant throughout the motion.... >oriDontal component of velocity at 7 velocity at C horiDontal component of

velocity at v1cos u v2cos(?*P - v1cos u v2sintaking first two terms, we get

cos uv1taking second two terms, we get


13/15

sin uv2=quaring and adding (i O (ii

cos2+ sin2 u2v12+ u2v22

1u2 1v12+ 1v2

2.

Q.C particle is proJected from a point at an angle and after time t is oserved to

have an elevation as seen from the point of proJection prove that the initial velocitywas (gt cos(2sin(- .

Ans.Let the particle e proJected from point ! with velocity u at an angle withthe horiDon !I. Let the particle e at 7 after time t. =uch that !7 E O I!7 .

;raw perpendicular 7C on !I from 7, then in !C7, 7C E sin O !C Ecos

:n time t, horiDontal O vertical distances transverse y the particle are !C O 7C.4ow the horiDontal O vertical components of the initial velocity are ucosO usin.The horiDontal distance covered y the particle in time t is !C

!C (ucostEcos ucos.t .....................(iThe vertical distance coverd y the particle in time t is 7Ch ut - (12 gt2

7C (usint - gt2

Esin usin.t - (12gt2....................(ii;ividing (i y (ii, we get(Ecos(Esin (ucos.tusin.t - (12get20cos2usin- gt0 2ucossin

2usincos- cossin0 gt cosu (gtcos2sin(- >ence the initial velocity (gtcos2sin(- 0.

Q!.C particle is proJected with a velocity 2ag. =o that it Just clears two walls ofequal heights a which are at a distance 2a from each other. =how that the latus

rectum of the traJectory is equal to 2a and the time of passing etween the walls is a

2ag

Ans!.Let C' and G4 e two walls of equal heights a at a distance 2a. CG '4 2a

4ow, we know that the velocity of a proJectile at any point in its path is equal to the

velocity of free fall from the directriI to that point.>ight of the ditrectriI aove thepoint of proJection

u22g u 2gh h u22g0


14/15


15/15

(u sin t - (12gt2u0 cos . t ....... KMsing (ii and (iii (2usin - gt2ucos 2usin - (u sin - v sin02u cos ....... KMsing (iv u sin + v sin2u cos u sin + (u coscos . sin02u cos ...... KMsing (i 12 tan + tan0>ence, 7roved

elementary statics and dynamics

Documents